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Turaev torsion and cohomology for closed 3–manifolds is known (see [Tur02] Chapter III), and below we give the analogue for 3–manifolds with boundary. We will give a definition of Turaev torsion based upon the definition of Reidemeister torsion. First, a brief review of Reidemeister torsion. Recall the (commutative) Reidemeister torsion of a finite complex X and a ring map ϕ : Z[H1(X)] → F , where F is a field, is an element ϕ τ (X) ∈ F/(±ϕ(Z[H1(X)]). Given an Euler structure e and a homology orientation ω (see [Tur02] for definitions), one can define the refined ϕ–torsion τ ϕ(X,e,ω) ∈ F , which is a unit if the ϕ–twisted ϕ complex C∗ (X) is acyclic, and is equal to 0 ∈ F if not. 1.1. The Turaev Torsion. If X is a finite connected CW–complex, the quotient ring (i.e. the ring obtained by localizing at the multi- plicative set of non-zerodivisors) of the integral group ring of a finitely generated abelian group splits as a direct sum of fields (see [Tur02] I.3.1 for details). This isomorphism provides ring homomorphisms from Z[H1(X)] to various fields. Specifically, if we denote by Q(H) the quotient ring of Z[H] where H = H1(X), we have the inclusion ≈ Z[H] ֒→ Q(H). There is an isomorphism Φ: Q(H) → Fi where i each Fi is a field and i ranges over a finite index set. This isomorphismL is defined, for example, in [Tur02], and is unique up to unique isomorphism (which, after re-ordering if necessary, will decompose along the ′ direct sum as a component-wise isomorphism Fi → Fi ) making the following diagram commute:

Fi i x; xx xx L xx xx Q(H) F FF FF FF F# ′ Fi . i L Then denote by ϕi the map Z[H] → Fi consisting of the inclusion to Q(H) followed by the natural projection to Fi. Then for any homology orientation ω and Euler structure e, we deﬁne the Turaev torsion τ(X,e,ω) by

τ(X,e,ω) = Φ−1 τ ϕi (X,e,ω) ∈ Q(H). i ! M TURAEV TORSION AND COHOMOLOGY DETERMINANTS 3

This deﬁnition does not depend on Φ (by the uniqueness of Φ). Henceforth, the symbol τ(X,e,ω) will refer to the Turaev torsion of (X,e,ω) unless otherwise speciﬁed. The symbol τ ϕ(X,e,ω) will still refer to ϕ–torsion. The set of Euler structures possesses a free, transitive H1(X)–action, and we will use −ω to represent the opposite homology orientation to ω. One may easily show that τ(X, he, ω)= hτ(X,e,ω) τ(X, e, −ω)= −τ(X,e,ω).

1.2. Turaev Torsion and Alexander Invariants. This section re- peats a result from [Tur02] II.1.7. Though this result is unproven in [Tur02], it is a simple exercise to prove it via the method of proof of [Tur02] Theorem II.1.2. We will use the result in the proof of Theo- rem 2.2. Henceforth, we will often need to strike a column from a matrix. We will use the notation A(r) for the matrix obtained by striking the rth column from a matrix A. Theorem 1.1 (Turaev). Suppose M is a 3–manifold with ∂M =6 ∅, χ(M)=0. Then there is an Euler structure e such that for any homology orientation ω, and for any 1 ≤ r ≤ m, m+r (1) τ(M,e,ω)(hr − 1)=(−1) τ0 det(∆(r)).

2. The Integral Cohomology Ring 2.1. Determinants. Let M be a 3–manifold with boundary ∂M =6 ∅, 1 and suppose χ(M) = 0. Also assume b1(M) ≥ 2 so that H (M,∂M) =6 1 1 1 0. Let fM denote the map H (M,∂M) × H (M) × H (M) −→ Z, defined by (a, b, c) 7→ ha ∪ b ∪ c, [M]i, where [M] is the fundamental class in H3(M,∂M) determined by the orientation and h·, ·i denotes evaluation pairing. This is alternate in the last two variables; ha ∪ b ∪ c, [M]i = −ha ∪ c ∪ b, [M]i. Our assumption χ(M) = 0 implies 1 that H (M,∂M) will have rank b1(M) − 1. There is a notion of a “determinant” of an alternate trilinear form (see [Tur02], chapter III for a discussion of the determinant of the obvious analogue of the above form when M is closed), but because of the difference in rank, we must have a new concept of determinant for a mapping such as the one above. The determinant of an alternate trilinear form on a free R– module is independent of basis up to squares of units of R, so if R = Z it is independent of basis. This will not be true of our determinant; 4 CHRISTOPHER TRUMAN however we will present a sign-refined version based on a choice of homology orientation. For our usage, this is not more of a choice than we would normally make; if we want sign-refined torsion, then we have already chosen a homology orientation, and if we do not care about the sign of the torsion, we can ignore the sign here as well. Let R be a commutative ring with unit, and let K, L be finitely generated free R modules of rank n and n − 1 respectively, where n ≥ 2. Let S = S(K∗) be the symmetric algebra on K∗, where K∗ = n ∗ n HomR(K, R). Note if {ai}i=1 is a basis of K with dual basis {ai }i=1 of ∗ ∗ ∗ ∗ ∗ K , then S = R[a1,...,an], the polynomial ring on a1,...,an, and the grading of S corresponds to the usual grading of a polynomial ring. n n−1 ∗ Now let {ai}i=1, {bj}j=1 be bases for K, L respectively, and let {ai } ∗ be the basis of K dual to the basis {ai}. Let f : L×K×K −→ R be an R–module homomorphism which is skew-symmetric in the two copies of K; i.e. for all y, z ∈ K, x ∈ L, f(x, y, z)= −f(x,z,y). Let g denote the associated homomorphism L × K −→ K∗ given by (g(x, y))(z) = f(x, y, z). Next we state a Lemma defining the determinant of f (d in the Lemma), but first we set some notation: [a′/a] ∈ R× is used to denote the determinant of the change of basis matrix from a to a′, and for a matrix A, we will let A(i) denote the matrix obtained by striking out the ith column as above.

Lemma 2.1. Let θ denote the (n − 1 × n) matrix over S whose i, jth entry θi,j, is obtained by θi,j = g(bi, aj). Then there is a unique d = d(f, a, b) ∈ Sn−2 such that for any 1 ≤ i ≤ n,

i ∗ (2) det θ(i)=(−1) ai d. For any other bases a′, b′ of K, L respectively, we have

(3) d(f, a′, b′) = [a′/a][b′/b]d(f, a, b).

∗ Proof. Let β denote the (n − 1 × n) matrix with βi,j = g(bi, aj)aj . The sum of the columns of β is zero; indeed, for any i, the ith entry (of the column vector obtained by summing the columns of β) is given by:

n n n ∗ ∗ ∗ βi,j = g(bi, aj)aj = f(bi, aj, ak)aj ak =0. j=1 j=1 X X j,kX=1 The last equality follows since the f term is anti-symmetric in j, k and the a term is symmetric. It is then a simple algebraic fact that since β is an n − 1 × n matrix whose columns sum to zero, (−1)i det β(i) is independent of i. TURAEV TORSION AND COHOMOLOGY DETERMINANTS 5

n−1 Let ti = det θ(i) ∈ S . It is clear that

∗ det(β(i)) = ti ak, Yk6=i thus for any i, p ≤ n, we have n i ∗ ∗ i ∗ ∗ (−1) tiap ak = (−1) det β(i)apai k=1 Y p ∗ ∗ = (−1) det β(p)ai ap n p ∗ ∗ = (−1) tpai ak. kY=1 ∗ Since the annihilators of ak in S are zero, we must have i ∗ p ∗ (−1) tiap =(−1) tpai . ∗ ∗ ∗ This means that ai divides tiap for all p, hence ai divides ti. Deﬁne si ∗ by ti = siai . Note i ∗ ∗ i ∗ p ∗ p ∗ ∗ (−1) siai ap =(−1) tiap =(−1) tpai =(−1) spapai ,

i i hence (−1) si is independent of i. Let d =(−1) si. By deﬁnition,

i i i ∗ ∗ (−1) det θ(i)=(−1) ti =(−1) siai = ai d. This proves (2). Now to prove the change of basis formula (3), we will ﬁrst show ′ ′ d(f, a , b) = [a /a]d(f, a, b). Let Si be the (n × n − 1) matrix obtained by inserting a row of zeroes into the (n − 1 × n − 1) identity matrix as the ith row. Then one may easily see for any (n − 1 × n) matrix A, the matrix A(i) (obtained by striking out the ith column) can also be + obtained as A(i) = ASi. Let Si denote the (n × n) matrix obtained by appending a column vector with a 1 in the ith entry and zeroes + otherwise on to the right of Si (i.e. Si is the identity matrix with th i the i column moved all the way to the right), and let A+ denote the (n × n) matrix obtained by appending a row vector with a 1 in the ith entry and zeroes otherwise on to the bottom of A. Note

+ n+i det(Si )=(−1) i n+i det(A+)=(−1) det(A(i)) hence i + det(A+Si ) = det(ASi) = det(A(i)). 6 CHRISTOPHER TRUMAN

′ ′ Now let (a /a) denote the usual change of basis matrix so that ai = n ′ (a /a)i,jaj. Now θi,j = g(bi, aj), so let j=1

P ′ ′ θi,j = g(bi, aj). One can easily show θ′ = θ · (a′/a)T, where T denotes transpose. Now we compute

i ′ T + i ′ T + det θ+(a /a) Si = det θ+ det (a /a) det Si ′ T i + = det (a /a) det θ+ det Si ′ i + = det (a /a) det θ+· Si ′ = [a /a] det(θ(i)) ′ i ∗ = [a /a](−1) ai d(f, a, b). To complete the proof, we multiply the argument of the determinant in the left-hand side of the above equation out. Let ei denote the row vector with a 1 in the ith position and zeroes otherwise, i.e. the ith th basis vector of a as expressed in the a–basis, and let ri denote the i ′ T th row of (a /a) and ci denote the i column. Then

i ′ T + θ ′ T T Si e det(θ+(a /a) Si ) = det ei (a /a) ( i )

h θ (a′/a)T(i) c i = det ei ( i ) ′ T θ(a /a) (i) θci = det ′ T ′ T (a /a) (i)i (a /a)i,i ′ T = (−1)n−i det θ(a /a) ri n−i θ′ = (−1) det ri n n−i n+k ′ T ′ = (−1) (−1) (a /a)i,k det(θ (k)) k=1 Xn n−i n+k ′ T k ′ ∗ ′ = (−1) (−1) (a /a)i,k(−1) (ak) d(f, a , b) k=1 Xn n−i n+k ∗ ′ ∗ k ′ ∗ ′ = (−1) (−1) (a /(a ) )i,k(−1) (ak) d(f, a , b) Xk=1 i ′ ∗ = (−1) d(f, a , b)ai . ′ ′ ∗ So d(f, a , b) − [a /a]d(f, a, b) annihilates ai for each i, hence is zero. The computation for a b change of basis is easier. Let b′ be another basis for L and let (b′/b) denote the b to b′ change of basis matrix. Let θ′ ′ ′ ′ ′ ′ denote the matrix g(bi, aj), then θ =(b /b)θ. So θ Si =(b /b)θSi, hence TURAEV TORSION AND COHOMOLOGY DETERMINANTS 7 det(θ′(i)) = [b′/b] det(θ(i)). This proves d(f, a, b′) = [b′/b]d(f, a, b), and completes the proof of (3).

2.1.1. The Sign Refined Determinant. In the case, R = Z, our determinant depends on the basis only by its sign. In this case, we can refine the determinant by a choice of orientation of the R–vector space (K ⊕ L) ⊗ R. Let ω be such a choice of orientation. Then define Detω(f) = det(f, a, b) where a, b are bases of K, L respectively such that the induced basis of (K ⊕ L) ⊗ R given by extension of scalars is positively oriented with respect to ω. Then Detω(f) is well defined, ′ ′ ′ ′ and for any bases a , b , we have det(f, a , b )= ± Detω(f) where the ± is chosen depending on whether a′, b′ induces a positively or negatively oriented basis of (K ⊕ L) ⊗ R with respect to ω. Note that for K = H1(M), L = H1(M,∂M) where M is a compact connected oriented 3–manifold with non-void boundary, a choice of homology orientation will determine an orientation for (K ⊕ L) ⊗ R. One simply says that a, b is a positively oriented basis for H1(M, R) ⊕ H1(M,∂M; R) with respect to a homology orientation ω if and only ∗ ∗ if {[pt], a1,...,an, b1 ∩ [M],...,bn−1 ∩ [M]} is a positively oriented ba- ∗ sis in homology (where a is the basis of H1(M, R) dual to a,[M] is the fundamental class determined by the orientation, and [pt] is the homology class of a point). This sign refinement is identical to refining Det by the paired volume form associated to ω, as defined below in (12). Also, note that this sign refined determinant only depends on the homology orientation, not the orientation of M.

2.2. Relationship to Torsion. We use the above to relate the torsion to the cohomology ring structure. Let T = Tors(H1(M)) denote the torsion subgroup of H1(M). Note that this is isomorphic to the torsion subgroup of H1(M,∂M), so we will also denote the torsion subgroup of H1(M,∂M) by T . Let G = H1(M)/T , let S(G) denote the graded symmetric algebra on G and let I denote the augmentation ideal in Z[H1(M)]. The filtration of Z[H1(M)] by powers of I determines an associated graded algebra A = Iℓ/Iℓ+1. Then there ℓ≥0 is an additive homomorphism q : S(G) −→ A Ldefined in [Tur02]. We repeat the definition here: The map h 7→ h − 1 mod I2 defines an 2 additive homomorphism H1(M) −→ I/I . This extends to a grading- preserving algebra homomorphism qH1(M) : S(H1(M)) −→ A. Any section s: G −→ H1(M) of the natural projection H1(M) −→ G induces e 8 CHRISTOPHER TRUMAN an algebra homomorphism s: S(G) −→ S(H1(M)); set

q = |T |qH1(M)s: S(G) −→ A. e Then q is grading preserving and is a Z–module homomorphism, and obviously satisﬁes the multiplicativee e formula q(a)q(b)= |T |q(ab). The map q does not depend on the choice of section s (see [Tur02]). We are now ready to state the main result of this section:

1 1 1 Theorem 2.2. Let fM : H (M,∂M) × H (M) × H (M) −→ Z be the Z–module homomorphism deﬁned by

fM (x, y, z)= hx ∪ y ∪ z, [M]i.

Let n = b1(M) ≥ 2, let I be the augmentation ideal of Z[H1(M)], and let e be any choice of Euler structure on M and ω be a homology orientation of M. Then τ(M,e,ω) ∈ In−2 and: n−1 n−2 n−1 (4) τ(M,e,ω) mod I = q(Detω(fM )) ∈ I /I . That τ(M,e,ω) ∈ In−2 is proved in [Tur02], Chapter II, the theorem is concerned with its image modulo In−1; this is the “leading term” of the torsion in the associated graded algebra A. This proof is generally the method of [Tur02] Theorem 2.2, though some adjustments must be made to incorporate the relative homology. Proof. The ﬁrst step is to arrange a handle decomposition coming from a C1 triangulation to be in a convenient form for comparing the torsion to the cohomology. First, we arrange our decomposition for M so that we have (0) 3– handles, (m − 1) 2–handles, (m) 1–handles, (1) 0–handle, and this is Poincar´edual to a relative handle decomposition for (M,∂M) with (0) 0–handles, (m−1) 1–handles, (m) 2–handles, (1) 3–handle. With these decompositions, we have the following cellular chain complexes:

m−1 m 0 C∗(M): 0 −−−→ Z −−−→ Z −−−→ Z

≈ ≈ ≈ ≈

 m−1 m  C∗(M,∂M): 0 ←−−− Z  ←−−− Z ←−−− Z. y y y 0 y We will refer to the handles as “honest” handles and “relative” handles; honest handles being from the decomposition of M and relative ones from the relative decomposition of (M,∂M). Later, we will ex- plicitly give the (m − 1 × m) matrix for ∂2 of the honest decomposition (i.e. the only nonzero boundary in the honest complex). TURAEV TORSION AND COHOMOLOGY DETERMINANTS 9

The core 0–cell of the honest 0–handle (of M) is a point, u, which we will say is positively oriented. At the same time we orient the relative 3–handle (of (M,∂M)) with the positive orientation given by the orientation of M. Extend the core 1–disks of the honest 1–handles to obtain loops in M based at u, representing x1,...,xm ∈ π1(M,u). Since sliding the ith honest 1–handle over the jth honest 1–handle replaces xi with xixj, and reversing orientation of the core 1–disk changes −1 replaces xi with xi , we may perform handle moves to assume that the images of the homology classes of the first n of the xi’s form a basis of G = H1(M)/T and the rest of the classes end up in T . For i = 1,...,m, let hi be the image of xi in H1(M) under the Hurewizc map, and hi = hi mod T . Thus h1,..., hn is a basis of G and hi = 1 1 ∗ ∗ for i > n. Denote the dual basis of H (M) by h1,...,hn, by definition, ∗ hhi , hji = δei,j, where h·, ·i is evaluatione pairing.f e We now arrange the relative 1–handles in a similar way; so that the imagese of the first n − 1 of them form a basis of H1(M,∂M)/T and the other m − n of them end up in the torsion subgroup. Let c denote the number of components of ∂M, we will now describe a method to geometrically realize the splitting of H1(M,∂M) as the image of c−1 H1(M) direct sum with Z generated by paths connecting boundary components. We will arrange for the first c − 1 of the relative handles to connect a given boundary component to the other components, and the rest of the handles to represent loops based at a point on that component. Since M is connected, the relative 1–skeleton is path connected, which means boundary component has at least one relative 1–handle with only one endpoint on that component. Choose a base point in one boundary component, call that component (∂M)0. By sliding relative handles over one another if necessary, we can arrange so that there is a relative 1–handle connecting (∂M)0 to each other boundary component. This is reasonably simple to do; given another boundary component, there is a path through the relative 1–skeleton connecting that component to (∂M)0. The first part of this path is a relative 1– handle with one endpoint in the boundary component we would like to connect to (∂M)0, so one simply slides the other end of the relative 1–handle along the path to (∂M)0, which is simply sliding it over other boundary components and other relative 1–handles, until it con- nects our boundary component to (∂M)0. Once we have fixed c − 1 relative handles doing this (one for each boundary component which is not (∂M)0), slide all of the other relative handles along the fixed c − 1 10 CHRISTOPHER TRUMAN handles to obtain relative handles whose cores generate the image of the fundamental group of M based at the chosen base point. To summarize, by handle moves, we arrange so that the first c − 1 relative handles are paths connecting a chosen boundary component to all of the other boundary components, and the other m − c relative handles have homology classes generating the image of H1(M) in H1(M,∂M). With these last handles, we may proceed as before in the discussion of honest handles; arrange so that the first n − c of these handles will give us the remaining free generators of H1(M,∂M)/T and the rest of them simply end up in T (again by sliding handles, since the only handles that we now need to slide represent loops all based at the same point). We will use similar notation, ki will denote the th homology class of the i handle and ki = ki mod T . We will denote 1 ∗ ∗ the dual basis of H (M,∂M) by k1,...,kn−1. As before, the ki’s for i ≤ n − 1 are generators of H1(M,∂Me)/T and for i > n − 1, ki = 1. ∗ e Also, as before, hki , kji = δi,j. The attaching maps for the honest 2–handles determine (upe to con- jugation) certain elementse r1,...,rm−1 of the free group F generated by x1,...,xm. We now have π1(M) presented by the generators x1,...,xm and the relations r1,...,rm−1. Now the cellular chain complex for M is in a particularly convenient form for our purposes. As usual, we use the notation ∂p to denote the boundary map from dimension p to dimension p − 1. Clearly ∂1 is given by the zero map. Let us denote the matrix of ∂2 by (vi,j) where 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ m. Now for 1 ≤ i ≤ n − 1, the core 2–disk of the ith honest 2–handle ∗ represents a cycle in C2(M) (in fact, its homology class is ki ∩ [M]). So it has boundary equal to zero, hence vi,j = 0 for i ≤ n − 1 and all j. We apply the same argument to the relative handles as follows: the jth relative 2–handle represents a homology class Poincarédual ∗ to hi ∩ [M], hence has boundary equal to zero, and vi,j = 0 for all i and j ≤ n. The result is that vi,j = 0 except for the bottom right hand (m − n × m − n) corner of the matrix; call this matrix v. This 0 0 tells us that ∂2 in the complex for M is given by ( 0 v ). This v is a square presentation matrix for the torsion group T , thus det(v)= ±|T |. Furthermore, r1,...,rn−1 ∈ [F, F ] since the first n − 1 honest 2–cells are cycles. Essentially, sliding handles and relative handles amount to doing integral row and column operations to reduce the matrix for ∂2 to this nice form. Consider the chain complex C∗(M) associated to the induced handle decomposition of the maximal abelian cover M of M. This is a free c c TURAEV TORSION AND COHOMOLOGY DETERMINANTS 11

Z[H1(M)]-chain complex with distinguished basis determined by lifts of handles of M. For an appropriate choice of these lifts, we have th ∂1 given by x 7→ x · w where w is a column of height m whose i entry is hi − 1. The map ∂2 is given by the Alexander-Fox matrix for the presentation hx1,...,xm|r1,...,rm−1i for an appropriate choice of the ri’s in their conjugacy classes. This is an (m − 1 × m) matrix th whose (i, j) entry is given by η(∂ri/∂xj ) where η is the projection Z[F ] −→ Z[π1(M,u)] −→ Z[H1(M)]. Let eN be an Euler structure determined by a fundamental family of cells which gives this “nice” cellular structure to M. Clearly the Z[H1(M)]–complex for M must augment to the Z–complex for M, hence aug(η(∂ri/∂xj )) = vi,j. From before, we know that cvi,j = 0 except in the lower right-hand corner,c so η(∂ri/∂xj ) ∈ I for i ≤ n − 1, j ≤ n. We claim for i ≤ n − 1, j ≤ n, n ∗ ∗ ∗ 2 (5) |T |η(∂ri/∂xj )= −|T | hki ∪ hj ∪ hp, [M]i(hp − 1) mod I . p=1 X Here I is the augmentation ideal. We will prove this by looking at η, the composition of η with the projection Z[H1(M)] −→ Z[G]. If we let J denote the augmentation ideal in Z[G], then it is enough to showe that for i ≤ n − 1, j ≤ n, n ∗ ∗ ∗ 2 (6) η(∂ri/∂xj)= − hki ∪ hj ∪ hp, [M]i(hp − 1) mod J . p=1 X e To provee (6), note that J/J 2 is isomorphic to the free abelian group G of rank n under the map g 7→ (g − 1) mod J 2, and is thus generated by h1 − 1,..., hn − 1. To be precise, for any g ∈ G, the expansion n ∗ hhp,gi g = hp gives ep=1 e Q e n ∗ 2 (7) g − 1= hhp,gi(hp − 1) mod J . p=1 X Also, for any α ∈ F, j ≤ n, e

∗ (8) aug(∂α/∂xj )= hhj , η(α)i.

Now ri ∈ [F, F ] gives an expansion ri = [αµ, βµ] a ﬁnite product of µ commutators in F . Then Q

η(∂ri/∂xj )= (η(αµ) − 1)η(∂βµ/∂xj )+(1 − η(βµ))η(∂αµ/∂xj ). µ X 12 CHRISTOPHER TRUMAN

Projecting to Z[G] we get

2 η(∂ri/∂xj) mod J = n ∗ ∗ ∗ ∗ (9)e hhp, η(αµ)ihhj , η(βµ)i − hhp, η(βµ)ihhj , η(αµ)i (hp − 1). p=1 µ ! X X e Now we consider the handlebody U ⊂ M formed by the (honest) 0–handle and the (honest) 1–handles. The boundary circle of the core th 2-cell of the i honest 2–handle lies in ∂U and represents ri. The expansion ri = [αµ, βµ] tells us that the circle is the image of the boundary µ ′ ′ ′ under a mapQ σi : Σi → U, where Σi is a surface which have meridians and longitudes with homology classes aµ, bµ respectively, where ′ ′ ′ (σi)∗(aµ)= η(αµ) and (σi)∗(bµ)= η(βµ) for all µ. Let Σi be Σi capped ′ with a disk; the map σi extends over the disk to a map σi : Σi → U by mapping the capping 2–disk to the core 2–cell of the ith honest handle. We may orient Σi so that the fundamental class (σi)∗([Σi]) is represented in H2(M) by the homology class of the core disk of the th ∗ i 2–handle, hence (σi)∗([Σi]) = ki ∩ [M]. Now for any 1-cohomology ′ classes ti, ti of Σi, we have

′ ′ ′ (10) hti ∪ ti, [Σi]i = hti, aµihti, bµi−hti, bµihti, aµi. µ X ∗ Pulling hj back to Σi we get a 1-cohomology class whose evaluations ∗ ∗ on the meridians and longitudes are hhj , η(αµ)i and hhj , η(βµ)i respectively. This proves (everything modJ 2)

n ∗ ∗ (11) η(∂ri/∂xj )= hhp ∪ hj , (σi)∗([Σi])i(hp − 1) p=1 X n e e ∗ ∗ ∗ = hhp ∪ hj ,ki ∩ [M]i(hp − 1) p=1 X n e ∗ ∗ ∗ = hhp ∪ hj ∪ ki , [M]i(hp − 1) p=1 X n e ∗ ∗ ∗ 2 = − hki ∪ hj ∪ hp, [M]i(hp − 1) mod J . p=1 X e This proves (6) which in turn proves (5). Recall by [Tur02] II.4.3, we have τ(M,e,ω) ∈ Z[H1(M)]. We have ar- ranged our handles so that h1 in particular has inﬁnite order in H1(M), TURAEV TORSION AND COHOMOLOGY DETERMINANTS 13 so by (1), we have m+1 (h1 − 1)τ(M, eN ,ω)=(−1) τ0 det(∆(1)).

Recall eN is chosen so that we may use (1). We now want to work out τ0. For now we work in a very specific homology basis: ∗ ∗ {[pt], h1,...,hn,k1 ∩ [M],...,kn−1 ∩ [M]}. Let ω denote the homology orientation defined by this basis, and let [ω/ω] denote the sign of ω with respect to ω. Later, when we do the 1 1 Det(ef) calculation, we will use the bases for H (M) and H (M,∂M) ∗ ∗ ∗ ∗ givene by {h1,...,hn} ande{k1,...,kn−1} respectively. Using the basis above (by which we defined ω), we compute |C∗(M)|+n(m−n) τ(C∗(M; R))=(−1) det v, e where v is defined as above. This is a quick calculation; we may choose our bases of the images of the boundary maps so that the dimension 2 and dimension 0 change of basis matrices are the identity matrices. Then the dimension 1 change of basis matrix will be the block 0 v matrix ( id 0 ), where id represents the (n × n) identity matrix. This has determinant (−1)n(m−n) det v. Another quick calculation gives m |C∗(M)| = (mn + m + n) mod 2. Hence τ0 = [ω/ω](−1) sign(det v). This gives m+m+1 (h1 − 1)τ(M, eN ,ω) = [ω/ω](−1) sign(dete v) det(∆(1)). Let a denote the submatrix of ∆ comprised of the first n − 1 rows and n columns; thus a is the matrixe whose i, j entry is given by η(∂ri/∂xj) for 1 ≤ i ≤ n − 1 and 1 ≤ j ≤ n. Let V denote the lower right hand (m−n×m−n) matrix η(∂ri/∂xj ) for n ≤ i ≤ m−1 and n+1 ≤ j ≤ m. Hence

(h1 − 1)τ(M, eN ,ω)= −[ω/ω]| det v| det a(1) = −[ω/ω]|T | det a(1) mod In. Define e n ∗ e∗ ∗ θi,j = hki ∪ hj ∪ hp, [M]ihp. p=1 X Then by (5), the entries of the matrix |T |a moduloe I2 can by obtained from the entries of the matrix |T |θ by replacing each hp by (hp − 1). From the definition of q above, we see n e (h1 − 1)τ(M, eN ,ω) mod I = −[ω/ω]q(det(θ(1))). By definition, i.e. (2), e det(θ(1)) = −[ω/ω]Detω(fM )h1.

e e 14 CHRISTOPHER TRUMAN

When we put this together, all of the signs will neatly cancel out, leaving

n (h1 − 1)τ(M, eN ,ω) mod I =(h1 − 1)q(Detω(fM )).

ℓ ℓ+1 ℓ ℓ+1 Then the map I /I deﬁned by x ∈ I /I maps to (h1 − 1)x ∈ ℓ≥0 Iℓ+1/Iℓ+2 is injective,L so

n−1 τ(M, eN ,ω) mod I = q(Detω(fM )).

But now recall τ(M,e,ω) only diﬀers from τ(M, eN ,ω) by multiplica- n−2 n−1 tion by an element of H1(M). They are both in I , so mod I they are equal. This completes the proof.

3. The Cohomology Ring Mod–r In this section, we will prove an analogous result to the one in Sec- tion 2 using cohomology modulo an integer r rather than integral cohomology. The integer r will have to be one such that the first cohomology group with Mod–r coefficients is a free Zr–module; for instance if r is prime. This will also imply that the first relative cohomology group is a free Zr–module, so we will still be able to compute a determinant as in Section 2, however will will need to refine that determinant slightly. To do so, we will first introduce the concept of a paired volume form, which will play a similar role to the square volume forms found in [Tur02], III.3 Before anything else, however, let us define the Mod–r torsion. This is defined when b1(M) ≥ 2 so that τ(M,e,ω) ∈ Z[H1(M)] for any e, ω. Then τ(M,e,ω; r) is the image of τ(M,e,ω) under the projection Z[H1(M)] → Zr[H1(M)] induced by the coefficient projection e1 e2 ek Z → Zr. Note that if r = p1 · p2 ··· pk where p1,...,pk are primes, then Zr[H1(M)] splits naturally as Z e1 [H1(M)] ⊕···⊕ Z ek [H1(M)] p1 pk e1 ek and τ(M,e,ω; r) splits as τ(M,e,ω; p1 )+ ··· + τ(M,e,ω; pk ), so un- derstanding Mod–r torsion when r is a power of a prime is sufficient to understand it for any r. One may also define the Mod–r torsion when b1(M) = 1 by using Turaev’s “polynomial part” [τ] of the torsion; see [Tur02], II.3. The- orem 3.3 is true in this case as well, and one can use the argument in [Tur02] Theorem III.4.3 when the first Betti number is 1 (the last paragraph of the proof).

3.1. Determinants. TURAEV TORSION AND COHOMOLOGY DETERMINANTS 15

3.1.1. Volume Forms. First we recall some definitions from [Tur02], III.3. If N is a finite rank free module over R, a commutative ring with 1, then a volume form ω on N is a map which assigns to each basis a of N a scalar ω(a) ∈ R such that ω(a) = [a/b]ω(b) for any bases a, b. A square volume form is a map Ω which also assigns a scalar to each basis, but the change of basis formula is Ω(a) = [a/b]2Ω(b). Naturally, the square of a volume form is a square volume form. This notion is useful when working with closed manifolds as in [Tur02], III.3, but we must use a slightly different form in the case of a nonempty boundary, though in the same spirit. If K, L are two finite rank free R–modules, then a paired volume form on K × L is a map µ from (ordered) pairs of bases of K and L to R such that µ(a′, b′) = [a′/a][b′/b]µ(a, b) where a, a′ are bases of K and b, b′ are bases of L. Note that the product of a volume form on K with a volume form on L is a paired volume form on K × L, so this notion is very similar to the notion of a square volume form. We say a paired volume form is non-degenerate if its image lies in the units of R, or equivalently if there is a basis a of K and a basis b of L so that µ(a, b) = 1. Note that we may easily construct a nondegenerate paired volume form given distinguished bases a, b of K, L respectively by assigning µ(a, b) = 1, and extending to other bases by the change of basis formula. Note the following properties of paired volume forms:

(1) If B : K × L → R is a bilinear form, where K and L are isomorphic R–modules, then µ(a, b) = det(Ba,b) is a paired volume form, where Ba,b is the matrix of B with respect to the bases a and b of K and L respectively, and µ is non-degenerate if and only if B is a nondegenerate form, i.e. if B induces an isomorphism K → HomR(L, R). (2) If K, L are free Z–modules of ﬁnite rank rK and rL respectively, and ω is an orientation on (K × L) ⊗ R, then there is a nondegenerate paired volume form µω on K×L such that µω(a, b)=

1 if the basis a1 ⊗ 1,...,arK ⊗ 1, b1 ⊗ 1,...,brL ⊗ 1 is positively oriented with respect to ω (and obviously µω assigns −1 to bases which are negatively oriented with respect to ω). (3) If 0 −→ K1 −→ K −→ K2 −→ 0 and 0 −→ L1 −→ L −→ L2 −→ 0 are short exact sequences of ﬁnite rank free R–modules and µ1,µ2 are paired volume forms on K1 × L1,K2 × L2, then there is an induced paired volume form on K ×L, which is nondegenerate if and only if µ1 and µ2 are both non-degenerate. To construct this, let ai, bi be bases of Ki, Li respectively. Then we can construct the bases a1a2 and b1b2 of K and L respectively 16 CHRISTOPHER TRUMAN

by concatenating the image of the basis a1 with a lift of the basis a2 in K, and similarly for b1b2 in L. Then for any bases a and b of K and L, define µ(a, b) = [a/a1a2][b/b1b2]µ1(a1, b1)µ2(a2, b2). (4) A non-degenerate paired volume form µ on K×L induces a nondegenerate paired volume form µ∗ on K∗ × L∗ ≈ (K × L)∗ = ∗ ∗ ∗ −1 ∗ HomR(K × L, R) by µ (a , b )=(µ(a, b)) where a is the basis of K∗ dual to a basis a of K, and similarly for b, b∗. (5) If φ: R → S is a surjection of rings, and µ is a nondegenerate paired volume form on the free R–modules K × L, then there is an induced paired volume form µφ on K ⊗S S ×L⊗S S given by µφ(a⊗1, b⊗1) = 1 if a, b are bases of K, L such that µ(a, b) = 1. 3.1.2. The Refined Determinant. Now given free R–modules K, L of finite ranks n and n−1 respectively (n ≥ 2), and given f : L×K×K → R an R–map as in Lemma 2.1, and given a paired volume form µ on K × L, we can construct the µ–refined determinant, Detµ(f), to be ∗ ∗ ∗ (12) Detµ(f)= µ (a , b )d(f, a, b) where d is defined as in Lemma 2.1. We can define this for any bases a, b of K, L respectively (and a∗, b∗ the dual bases as usual), but by the properties of d and µ, this is independent of the chosen bases. Note that this will simply be the determinant taken with respect to any bases a, b with µ(a, b) = 1 if such bases exist. 3.1.3. Constructing Paired Volume Forms. We now construct a paired volume form which we will later use to refine the determinant for mod–r cohomology. We will construct this form in pieces, and assemble them via methods enumerated above. Let H,H′ be finite abelian groups which are isomorphic, though we will not fix a particular isomorphism. (These groups will appear later as the torsion groups Tors(H1(M)) and Tors(H1(M,∂M))). Let p ≥ 2 be a prime integer dividing |H|. Let r = ps for some s ≥ 1 such that H/r is a direct sum of copies of Zr, so that we can think of H/r as a finite rank free Zr–module (and similarly for H′/r, since H,H′ are isomorphic). We will now show how to construct a paired volume form on H/r × H′/r from a bilinear form L: H × H′ → Q/Z. First, we repeat some definitions from [Tur02]. Let H(p) be the subgroup of H consisting of all elements annihilated ′ by a power of p (similarly for H(p)). A sequence h = (h1,...,hn) of nonzero elements of H(p) is a pseudo-basis if H(p) is a direct sum of the cyclic subgroups generated by h1,...,hn and the order of hi in H is less than or equal to the order of hj for i ≤ j. In other words, if the si order of hi is p , with si ≥ 1, then s1 ≤ s2 ≤···≤ sn. This sequence (s1,...,sn) is determined by H(p) and does not depend on the choice TURAEV TORSION AND COHOMOLOGY DETERMINANTS 17

k of pseudo-basis h. Note s ≤ s1 since if we have a summand of order p for k

(14) det(xi · yj) = [x/h][y/k] det(hi · kj). By symmetry, to prove (14) we need only show

(15) det(xi · kj) = [x/h] det(hi · kj).

Now x is a pseudo-basis for H(p), so the order of xi is equal to the order of hi for each i. It is clear that if x is just a permutation of h (the permutation can only permute elements of the same order), then the basis x of H/r is the same permutation of the basis h, and then (15) is clear. So now, we may assume that each xi generates the same cyclic subgroup of H(p) as the corresponding hi. Then for each i, there 18 CHRISTOPHER TRUMAN

si s is some ci ∈ Z, with ci coprime to p , hence coprime to r = p (in fact, coprime to p), with xi = cihi. But then xi · kj = (ci (mod r))hi · kj, so det(xi · kj)= i(ci (mod r)) det(hi · kj). But clearly [x/h] = i ci (mod r), so the proof of (15) is completed, and (14) follows. Q r Q Now if L is nondegenerate, then to show that µL is nondegenerate, × we must simply show that det(hi ·kj) ∈ Zr for any pseudo-bases h, k of ′ H(p),H(p) respectively. L nondegenerate means that the map induced ′ by L, L: H → HomZ(H , Q/Z), is a bijection. Then, in particular, the ′ restriction of L to H(p) is also bijective on its image HomZ(H(p), Q/Z). e ′ This means, for k any pseudo-basis of H(p), for each kj there is an −sj xj ∈ H(p) witheL(xi,kj)= δi,jp , i.e. xi · kj = δi,j. Then (15) gives us × ′ our result, that det(hi · kj) ∈ Zr for any pseudo-bases h, k of H(p),H(p) respectively. 3.1.4. The Q/Z linking form. We now construct a linking form on

Tors(H1(M)) × Tors(H1(M,∂M)). We use a slightly diﬀerent construction from the one in [Tur02], though one may easily verify that the end results are the same. From the Universal Coeﬃcient Theorem, there is an exact sequence

0 → H2(M) ⊗ Q/Z → H2(M; Q/Z) → Tor(H1(M), Q/Z) → 0 but Tors(H1(M)) is canonically isomorphic to Tor(H1(M), Q/Z) by

Tors(H1(M)) ≈ Tors(H1(M)) ⊗ Z

≈ Tor(Tors(H1(M)), Q/Z)

≈ Tor(H1(M), Q/Z) With this in mind, our exact sequence becomes

0 → H2(M) ⊗ Q/Z → H2(M; Q/Z) → Tors(H1(M)) → 0.

Now choose elements a ∈ Tors(H1(M)) and b ∈ Tors(H1(M,∂M)); we want to define their linking LM (a, b) ∈ Q/Z. So choose a ∈ 1 H2(M; Q/Z) mapping to a, then let α ∈ H (M,∂M; Q/Z) be Poincaré dual to a, i.e. α ∩ [M] = a. Then we define LM (a, b) = hα, bi ∈ Q/Z. One may show that starting with the sequence for H2(M,∂M; Q/Z) instead of the sequence for H2(M; Q/Z) yields the same form. 3.1.5. Constructing the Paired Volume Form for Cohomology. Let ω be a homology orientation. Then we have split exact sequences

0 → Tors(H1(M)) → H1(M) → H1(M)/ Tors(H1(M)) → 0,

0 → Tors(H1(M,∂M)) → H1(M,∂M) → H1(M,∂M)/ Tors(H1(M,∂M)) → 0. TURAEV TORSION AND COHOMOLOGY DETERMINANTS 19

Both of these sequences split, so they also split modulo r, and note

H1(M)/r ≈ H1(M; Zr) and H1(M,∂M)/r ≈ H1(M,∂M; Zr). The homology orientation induces a nondegenerate paired volume form on the free Z–module

H1(M)/ Tors(H1(M)) × H1(M,∂M)/ Tors(H1(M,∂M)) which induces a nondegenerate paired volume form on

(H1(M)/ Tors(H1(M)))/r × (H1(M,∂M)/ Tors(H1(M,∂M)))/r. Above, we showed how to construct a nondegenerate paired volume form (induced by the Q/Z–linking form) on

Tors(H1(M))/r × Tors(H1(M,∂M))/r. We may combine these to give a nondegenerate paired volume form on H1(M; Zr) × H1(M,∂M; Zr), which in turn gives us a nondegenerate paired volume form on the duals. We will denote the canonical Mod–r r paired volume form by on H1(M; Zr) × H1(M,∂M; Zr) by µM and the r 1 1 refined determinant of the form fM on H (M,∂M; Zr) × H (M; Zr) × 1 r H (M; Zr) by Detr(fM ). 3.2. Relationship to Torsion. We will now let I denote the augmentation ideal of Zr[H1(M)] instead of the augmentation ideal of Z[H1(M)] (the augmentation ideal of Zr[H1(M)] is the image of the augmentation ideal of Z[H1(M)] under the map induced by the coefficient projection Z → Zr). We now recall a definition from [Tur02]- ℓ ℓ+1 we define qr : S(H1(M)/r) → I /I by ℓ≥0

Lℓ ℓ+1 (16) qr(g1,...,gℓ)= (gi − 1) (mod I ) i=1 Y where gi is a lift of gi to H1(M) (thee proof that this is independent of the lift is in [Tur02]). Beforee we state the main theorem, we brieﬂy discuss Mod–r surfaces. In particular, we give equivalent equations to (10). An equivalent definition of Mod–r surfaces can be found in [Tur02] Section XII.3.

3.2.1. Mod–r surfaces. Let G(M, N ; r) be a group with generators αµ, βµ,γν where µ, ν run over ﬁnite indexing sets M, N respectively, r with a single relator ρ = [αµ, βµ] γν. Let X(M, N ; r) be a con- µ ν nected CW–complex withQ a single 0–cell,Q 1–cells aµ, bµ,cν (so that we 20 CHRISTOPHER TRUMAN can consider π1(X(M, N ; r)) to be generated by αµ, βµ,γν), and a single 2–cell attached along ρ, so it is obvious that

π1(X(M, N ; r)) ≈ G(M, N ; r).

Then H2(X(M, N ; r); Zr) ≈ Zr, so let [X(M, N ; r)] be the generator of H2(X(M, N ; r); Zr) given by the homology class of the two cell ′ 1 (whose boundary is zero modulo r). Now if t, t ∈ H (X(M, N ; r); Zr), then let us compute ′ ′ ht ∪ t , [X(M, N ; r)]i = ε∗ ((t ∪ t ) ∩ [X(M, N ; r)]) where ε∗ : H0(X(M, N ; r); Zr) → Zr is simply augmentation, [pt] 7→ 1. Let aµ, bµ,cν ∈ H1(X(M, N ; r); Zr) be the homology classes modulo r of αµ, βµ,γν respectively. We now claim

′ 1 Lemma 3.2. Let t, t ∈ H (X(M, N ; r); Zr). If r is odd, then

′ ′ ′ (17) ht ∪ t , [X(M, N ; r)]i = ht, aµiht , bµi−ht, bµiht , aµi. µ X If r is even, then (18) ′ ′ ′ r ′ ht ∪ t , [X(M, N ; r)]i = ht,aµiht ,bµi − ht,bµiht ,aµi + ht,cν iht ,cν i. 2 µ ν X X ∗ ∗ ∗ 1 Proof. If we let aµ, bµ,cν ∈ H (X(M, N ; r); Zr) be dual to aµ, bµ,cν un- ∗ ∗ ∗ ∗ der h·, ·i, then 1 = haµ ∪ bµ, [X(M, N ; r)]i = −hbµ ∪ aµ, [X(M, N ; r)]i. Clearly cν ∪ cν is 2–torsion for any r, and one can also show that all other cup products are zero (this follows from induction and a relatively simple Mayer–Vietoris argument). So the claim for r odd is completed. By the same Mayer–Vietoris argument, for even r, we only need to show the statement for M empty, and N only having one element, i.e. for even r, and a CW complex X with one 0–cell, one 1–cell c, and one 2– 2 r cell with boundary r ·c, we need to show hc , [X]i = 2 . But this follows from simply noting that X is the 2–skeleton of a K(Zr, 1). A more complete proof may be found in [Hat02] Chapter 3, Example 3.9. We are now ready to state the main theorem of this section.

Theorem 3.3. Let r be a power of a prime such that H1(M)/r = H1(M; Zr) is a free Zr–module of rank b ≥ 2. Let

T = | Tors(H1(M))|/r. Then for any Euler structure e and homology orientation ω, τ(M,e,ω; r) ∈ Ib−2, TURAEV TORSION AND COHOMOLOGY DETERMINANTS 21 and r b−1 (19) τ(M,e,ω; r)= T · qr(Detr(fM )) (mod I ). As in Theorem 2.2, that τ(M,e,ω; r) ∈ Ib−2 is proved in [Tur02] II.4.4, we are more concerned with the residue class modulo Ib−1. Proof. The proof is similar to the proof of Theorem 2.2, and again is the method of the proof of [Tur02] Theorem 4.3 with modiﬁcations to apply it to manifolds with nonvoid boundary. Suppose r = ps, where p ≥ 2 is prime and s ≥ 1. Let n = b1(M). Then H1(M) n ′ n−1 splits as Z × (Tors(H1(M)))(p) × H and H1(M,∂M) splits as Z × ′′ (Tors(H1(M,∂M)))(p) ×H where the subscript of (p) denotes the maximal subgroup of a ﬁnite group whose order is a power of p and H′,H′′ are (isomorphic) subgroups of Tors(H1(M)) and Tors(H1(M,∂M)) respectively with |H′| = |H′′| = T . We again choose a handle decomposition of M and the dual relative handle decomposition of (M,∂M) with 1 honest 0–handle, m honest 1–handles, m − 1 honest 2–handles, and no other handles, where m ≥ b ≥ n. Let x1,...,xm ∈ π1(M) be the generators of π1(M) (based at the 0–cell) given by the core 1– cells of the honest 1–handles, and let h1,...hm denote their homology classes. Let k1,...,km−1 denote the classes in H1(M,∂M) of the core cells of the relative 1–handles, and let r1,...,rm−1 be the relators in F = hx1,...,xmi given by the attaching maps of the honest 2–cells. Now, as in the proof of Theorem 2.2, we want to rearrange handles for a more convenient decomposition. As in the proof of Theorem 2.2, we can arrange the handle decomposition (by sliding handles) so that h1,...,hn are generators modulo Tors(H1(M)) and hn+1 ...,hm ∈ Tors(H1(M)). We can also arrange for hn+1,...,hb to be a pseudo-basis of (Tors(H1(M)))(p) by essentially the same method of sliding handles. The last handles then have ′ s1 sb−n the property hb+1,...,hm ∈ H . Let p ,...,p be the orders of hn+1,...,hb respectively, and we may assume s1 ≤ s2 ≤···≤ sb−n ≤ s. Now we will denote by h the projection of h ∈ H1(M) to H1(M)/r, then h1,..., hb is a basis for H1(M)/r over Zr and hi = 1 for i > b. ∗ 1 e ∗ Let hi ∈ H (M; Zr) for i ≤ b such that hhi , hji = δi,j. th Leteki denotee the class in H1(M,∂M) of the i relativee handle, using the methods in the proof of Theorem 2.2 and thee methods above, we can arrange so that k1,...,kn−1 are generators modulo Tors(H1(M,∂M)), kn,...,kb−1 form a pseudo-basis of (Tors(H1(M,∂M)))(p) (they also s1 sb−n ′′ have orders p ,...,p ) and kb,...,km−1 ∈ T . This means, using k to denote projection of k ∈ H1(M,∂M) to H1(M,∂M)/r, that k1,..., kb−1 is a basis for H1(M,∂M)/r over Zr and ki = 1 for i > b−1. e e e e 22 CHRISTOPHER TRUMAN

∗ 1 As above, let ki ∈ H (M,∂M; Zr) for i ≤ b − 1 be such that hki, kji = δi,j. The matrix for the boundary map from dimension two to dimensione 0 0 one in C∗(M) decomposes, as in the proof of Theorem 2.2, as ( 0 v ), where v is a square presentation matrix for Tors(H1(M)). Using the handle decomposition above, v can be split as the direct sum of a diagonal matrix (with ps1 ,...,psb−n along the diagonal) and a square matrix v′ which is a presentation matrix for H′ (and its transpose a presentation matrix for H′′), hence det(v′) = ±T . The diagonal submatrix of v (consisting of powers of p) arises as follows: for n +1 ≤ i ≤ b, hi si−n si−n has order p (according to the above argument); in fact, p hi is the boundary of the 2–cell transverse to ki. This 2–cell has boundary zero in Q/Z, and its homology class in H2(M; Q/Z) is Poincarédual to ∗ 1 the class of ki in H (M,∂M; Q/Z), which is dual under evaluation to the class of ki in Tors(H1(M,∂M)). This process is the precise process used in the construction of the linking pairing, first lifting an element of Tors(H1(M)) to H2(M; Q/Z) and then using Poincaréduality to get an element dual (under evaluation) to an element of Tors(H1(M,∂M)). The fact that this matrix is diagonal with psi−n running down the diagonal means that for n +1 ≤ i ≤ b and n ≤ j ≤ b − 1,

(20) (hi · kj)= δi,j.

Now as in the proof of Theorem 2.2, r1,...,rn−1 ∈ [F, F ], and the above argument shows rn,...,rb−1 can each be expanded as ri = r [αµ, βµ] γν, so we need to use Lemma 3.2. Henceforth, we will µ∈Mi ν∈Ni suppressQ the MQ i, Ni notation for simplicity. Now let pr: Z[H1(M)] → Zr[H1(M)] be coeﬃcient projection, let η : Z[F ] → Z[H1(M)] be induced by the projection F → H1(M) (through π1(M)), and let p: Zr[H1(M)] → Zr[H1(M)/r] be induced by H1(M) → H1(M)/r. Finally, we will also denote by ηr = p ◦ pr ◦ η. We now prove the analogue of (5) which is, for i ≤ b − 1, j ≤ b b r ∗ ∗ ∗ 2 (21) (pr ◦ η)(∂ri/∂xj )= − fM (ki , hj , hp)(hp − 1) (mod I ). p=1 X We will prove (21) by proving the analogue of (6), which is b r ∗ ∗ ∗ 2 (22) ηr(∂ri/∂xj )= − fM (ki , hj , hp)(hp − 1) (mod J ). p=1 X Note (21) follows from (22) since p induces ane isomorphism 2 2 Zr[H1(M)]/I → Zr[H1(M)/r]/J TURAEV TORSION AND COHOMOLOGY DETERMINANTS 23

(where J is the augmentation ideal in Zr[H1(M)/r]). This follows from r 2 r noting for any h ∈ H1(M), h − 1 ∈ I since (h − 1)=(h − 1)(1+ h + ···+hr−1), and (1+h+···+hr−1)=(h−1)+(h2 −1)+···+(hr−1 −1) in Zr[H1(M)]. To prove (22), we need to note that (7) and (8) can be used here mutatis mutandis; indeed, for c ∈ H1(M)/r, we may use the same formula as (7) with slightly diﬀerent meaning to the symbols

b ∗ 2 (23) c − 1= hhp,ci(hp − 1) (mod J ). p=1 X e Also, for any α ∈ F, j ≤ b, if we let augr denote aug ◦ p ◦ pr ◦ η, augr : Z[F ] → Zr,

∗ (24) augr(∂α/∂xj )= hhj , ηr(α)i.

For 1 ≤ i ≤ n − 1, we may compute ∂ri/∂xj by

η(∂ri/∂xj )= (η(αµ) − 1)η(∂βµ/∂xj )+(1 − η(βµ))η(∂αµ/∂xj ). µ X

For n ≤ i ≤ b − 1, we must add a term for the γν’s

η(∂ri/∂xj )= (η(αµ) − 1)η(∂βµ/∂xj)+(1 − η(βµ))η(∂αµ/∂xj ) µ X r−1 (25) + η(∂γν /∂xj)(1 + γν + ··· + γν ) . ν X 2 For any r, any c ∈ H1(M), working modulo I ,

r−1 r−1 cℓ = (1+(c − 1))ℓ Xℓ=0 Xℓ=0 r−1 ℓ ℓ = 1ℓ−s(c − 1)s s s=0 Xℓ=0 X r−1 = 1+ ℓ(c − 1) (mod I2) Xℓ=0 r−1 = ℓ(c − 1) Xℓ=0 =(c − 1)r(r − 1)/2. 24 CHRISTOPHER TRUMAN

2 So if r is odd, then each extra γν term is zero modulo I in (25), and if r is even (in our case, a power of two), then for each ν (applying (23)), r 1+ η (γ )+ ··· + η (γ )r−1 = − (η (γ ) − 1) r ν r ν 2 r ν r b = − hh∗, η (γ )i(h − 1) (mod J 2) 2 p r ν p p=1 X r b e (26) = hh∗, η (γ )i(h − 1) (mod J 2). 2 p r ν p p=1 X er r The last line follows since in Zr for an even r, − 2 = 2 . Now, using maps from Mod–r surfaces (i.e. Lemma 3.2) instead of maps from surfaces, we can use the proof from Theorem 2.2 since (9) holds for odd r, so (11) holds for odd r, proving (22) for odd r. For even r,(9) holds with an additional term following from (24) and (26). Speciﬁcally, for an even r, 2 ηr(∂ri/∂xj) mod J = b ∗ ∗ ∗ ∗ hhp, ηr(αµ)ihhj , ηr(βµ)i−hhp, ηr(βµ)ihhj , ηr(αµ)i p=1 µ X X r + hh∗, η (γ )ihh∗, η (γ )i (h − 1). 2 j r ν p r ν p ν ! X This term also occurs in (11) for even r bye Lemma 3.2, so (22) holds for even r as well, hence (21) holds for all r. If we let a be the submatrix of (pr ◦ η)(∂ri/∂xj ) consisting of the b × b − 1 upper left submatrix, ′ b (h1 − 1)τ(M,e,ω; r)= | det(v )| det(a(1)) = T det(a(1)) mod I .

Now by (21), computing det(a(1)) is simply computing qr(det(Θ(1))) b r ∗ ∗ ∗ where Θi,j = fM (ki , hj , hp)hp. p=1 P e r ∗ ∗ det(Θ(1)) = −h1d(fM , h ,k ). r ∗ ∗ From here, we may follow the proof from Theorem 2.2, since µM (h ,k ) will simply be a sign just as in Theoreme 2.2, since the linking form of the pseudo-bases is equal to the identity matrix, hence has determinant one. This follows from equation (20) which gives for n +1 ≤ i ≤ b and n ≤ j ≤ b − 1,

det(hi · kj) = det(δi,j)=1. TURAEV TORSION AND COHOMOLOGY DETERMINANTS 25

4. Integral Massey Products In this section, we give a generalization of Theorem 2.2 where we use Massey products rather than the cohomology ring. The results of this section are similar to results in Chapter XII Section 2 of [Tur02] for closed manifolds. Note that Massey products are related to Milnor’s µ¯– invariants for links in S3, much like cup products are related to linking numbers (see [Por80] and [Tur76]). 4.1. Determinants. First we obtain a new determinant. Let R be a commutative ring with 1, and let K, L be free R–modules of rank n,n − 1 respectively, with n ≥ 2 and let S = S(K∗), the symmetric algebra on the dual of K, as in Lemma 2.1. Let f : L × Km+1 → R be an R–map, with m ≥ 1. Define g : L × K → S by n ∗ ∗ g(x, y)= f(x, y, ai1 ,...,aim )ai1 ··· aim ∈ S i1,...,i =1 Xm n ∗ where {ai}i=1 is a basis for K and {ai } is its dual basis. This definition for g looks dependent on the basis chosen, however one can easily show that it is not. Let f0 : L → S be defined by n ∗ ∗ f0(x)= f(x, ai1 ,...,aim+1 )ai1 ··· aim+1 ∈ S. i1,...,i +1=1 Xm Again, f0 does not depend on the chosen basis, by precisely the same argument. Then we have the following lemma:

Lemma 4.1. Suppose f0 = 0. Let a = {ai}, b = {bj} be bases of K, L respectively, and let θ be the (n − 1 × n) matrix over S deﬁned by m(n−1)−1 θi,j = g(bi, aj). Then there exists a unique d = d(f, a, b) ∈ S such that i ∗ (27) det(θ(i))=(−1) ai d. Furthermore, if a′, b′ are other bases for K, L respectively, then (28) d(f, a′, b′) = [a′/a][b′/b]d(f, a, b).

Proof. This is very similar to the proof of Lemma 2.1. Let β be the ∗ matrix over S given by βi,j = g(bi, aj)aj . Then the sum of the columns n th of β is zero; the i entry in that sum is βi,j = f0(bi) = 0 since our j=1 P 26 CHRISTOPHER TRUMAN assumption is f0 = 0. Now the same argument as given in Lemma 2.1 to prove (2) completes the proof of (27), and the argument given to prove (3) can be used to prove (28). Note that as before, over Z the determinant is well defined up to sign, and that one may also sign-refine this determinant to remove the sign dependence. We may also define the condition that f is “alternate” in the K variables; let f0 : L × K → R be the R–map given by f0(x, a) = m+1times f(x, a, a, . . . , a). Then f0(x) = 0 for all x clearly implies f0(x, a)=0 for all x ∈ L, a ∈ K. The converse is also true provided that every polynomialz }| over{ R which only takes on zero values has all zero coefficients. This is true, for example, if R is infinite with no zero-divisors; in particular for R = Z. The rest of the argument is very similar to the argument in [Tur02], section XII.2. Let M be a 3–manifold with nonempty boundary, and 1 for u1,u2,...,uk ∈ H (M), let hu1,...,uki denote the Massey product 2 of u1,...,uk as a subset of H (M) (note in general this set may well be empty). See [Kra66] and [Fen83] for definitions and properties of Massey products. Now assume that m ≥ 1 is an integer such that 1 (∗)m: for every u1,...,uk ∈ H (M) with k ≤ m, hu1,...,uki =0

Here hu1,...,uki = 0 means that hu1,...,uki consists of the single ele- 2 ment 0 ∈ H (M). This condition guarantees that for any u1,...um+1 ∈ 1 H (M), the set hu1,...,umi consists of a single element; see [Fen83] Lemma 6.2.7. Deﬁne a Z–map f : H1(M,∂M) × (H1(M))m+1 → Z by m f(v,u1,...,um+1)=(−1) hv ∪hu1,...,um+1i, [M]i . The outermost h, i is used to denote the evaluation pairing.

Lemma 4.2. f0 = 0, so f has a well-deﬁned determinant (with the sign reﬁnement as above).

For m = 1, condition (∗)m is void, and in fact the Massey product hu1,u2i = −u1 ∪ u2, so this reduces to Lemma 2.1. Proof. By the argument above, we only need to show that f is alternate. But this follows from [Kra66] Theorem 15, which gives that for any element a ∈ H1(M), the m + 1 times Massey product of a m+1times with itself, ha,...,ai, lies in Tors(H2(M)), hence cupping with an element of H1(M,∂M) will give an element of Tors(H3(M,∂M)), which is null. z }| { TURAEV TORSION AND COHOMOLOGY DETERMINANTS 27

We will call this determinant Det(f), or if we care to introduce the sign-refined version with a homology orientation ω, Detω(f). Since the change of basis formula (28) is identical to the change of basis formula (3), the sign refinement by homology orientation is the same. 4.2. Relationship to Torsion. Theorem 4.3. Let M be a compact connected oriented 3–manifold with ∂M =6 ∅, χ(M)=0, n = b1(M) ≥ 2, and satisfying condition (∗)m for some m ≥ 1. Let e be an Euler structure on M, let ω be a homology orientation, and let q be defined as in Section 2. Define the form f as above. Then τ(M,e,ω) ∈ Im(n−1)−1 and m(n−1) m(n−1)−1 m(n−1) (29) τ(M,e,ω) mod I = q(Detω(f)) ∈ I /I . Proof. This proof is very much like the one in Section 2. In place of (4), we may use [Tur76] Theorem D, which gives the second line of the following string of equalities (all of which are modJ m+1) n m+1 η(∂ri/∂xj )= X aug(η(∂ ri/∂xi1 ...∂xim ∂xj ))(hi1 − 1) · · · (him − 1) e i1,...,im=1 e f e n ∗ ∗ ∗ = X ˙hhi1 ,...,him , hj i, (−[Σi])¸ (hi1 − 1) · · · (him − 1) i1,...,im=1 e e n ∗ ∗ ∗ ∗ = X ˙hhi1 ,...,him , hj i, (−ki ∩ [M])¸ (hi1 − 1) · · · (him − 1) i1,...,im=1 e e n ∗ ∗ ∗ ∗ = X − ˙ki ∪hhi1 ,...,him , hj i, [M]¸ (hi1 − 1) · · · (him − 1) i1,...,im=1 e e n m+1 ∗ ∗ ∗ ∗ (30) = X (−1) ˙ki ∪hhj , him ,...,hi1 i, [M]¸ (hi1 − 1) · · · (him − 1) i1,...,im=1 e e n m+1 ∗ ∗ ∗ ∗ = X (−1) ˙ki ∪hhj , hi1 ,...,him i, [M]¸ (hi1 − 1) · · · (him − 1) i1,...,im=1 e e n ∗ ∗ ∗ ∗ = X −f(ki , hj , hi1 ,...,him )(hi1 − 1) · · · (him − 1). i1,...,im=1 e e The line marked (30) follows from [Kra66] Theorem 8, and the next line is by symmetry. From here, the proof is identical to the proof of Theorem 2.2 after (11).

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Mathematics Department, University of Maryland, College Park, MD 20742 USA E-mail address: [email protected] URL: http://www.math.umd.edu/~cbtruman/